First slide

Thermal expansion

Question

A metal metre scale gives correct measurement at 0⁰C. It is generally used at a temperature of 40⁰C. The correction to be made for every metre is (α = 10^-6/⁰C)

Easy

A

4 × 10^–5 m
B

4 × 10^–5 m to be added
C

4 × 10^–5 m must be deducted
D

None of the above.

Solution

$\large l=1m,\;t_1=0^0C,\;t_2=40^0C;\;\alpha =10^{-6}/^0C$
we have $\large \alpha =\frac {\Delta l}{l\Delta t}=10^{-6}$
$\large 10^{-6}=\frac {\Delta l}{(1)(40-0)}\Rightarrow \Delta l=40\times 10^{-6}$
$\large =4\times 10^{-5}m(to\;be\;added)$

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